Suppose we view $S^3\subset C^2$. Then for coprime integers $p,q$ we define the lens space by $M_{p,q}=S^3/\sim$ where $(z_1,z_2)\sim(z_1e^\frac{2\pi i}{p},z_2e^\frac{2\pi iq}{p})$. I want to show that this is a smooth 3-manifold and that $M_{p,q}$ is diffeomorphic to $M_{p,q'}$ if $q'\equiv \pm q^{\pm 1}\pmod p$.
My thoughts were that we consider $(z_0,z_1)\in S^3$. Then $|z_0|^2+|z_1|^2=1$ and we can represent $z_0$ and $z_1$ as $r_0e^{i\theta_0}$ and $r_1e^{i\theta_1}$ respectively. Then we have $r_0^2+r_1^2=1$ so we have that $r_0=(1-r_1^2)^\frac{1}{2}$. This allows us to associate a real triple $(r_1,\theta_0,\theta_1)\in\mathbb{R}^3$ from any complex tuple $(z_0,z_1)\in\mathbb{C}^2$. Note that we have the constraints that $0\leq r_1\leq 1$, and $0\leq \theta_0,\theta_1\leq 2\pi$. Using this we can probably construct a smooth atlas but I am not exactly sure how to go about it. I am assuming we have to include the facts that $p$ and $q$ are coprime. One we have this atlas, I think that a diffeomorphism that makes sense to me would be $F:M_{p,q}\to M_{p,q'}$ defined by $F([z_1:z_2])=[z_1:z_2]$ but without the atlas theres no way for me to check that it is smooth given my current knowledge in the subject.